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3 years ago

Explain what WHMIS is

Chemistry

1 answer:

White raven [17]3 years ago

7 0

Workplace Hazardous Materials Information System is the answer to this question. Hope it helps :)

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What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m

r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.

4 0

3 years ago

How many moles are there in 7.5 L Of H2

True [87]

So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
$7.5 \div 22.4$
that 's what I think

3 0

3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0

3 years ago

What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use

ch4aika [34]

Explanation:

The given data is as follows.

$M_{1}$ = 10 mM = $10 \times 10^{-3}$ M

$V_{1}$ = 750 ml, $V_{2}$ = 5 ml

$M_{2}$ = ?

Therefore, calculate the molarity of given NaCl stock as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml$

$M_{2}$ = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.

7 0

3 years ago

Balance the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method.

LuckyWell [14K]

First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.

Mg --> Mg+2

<span>Hg+2 --> Hg+1
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation

$Mg$ ⇒ $Mg^{+2} + 2e^{-1}$

and

$2Hg^{+2} + 2e^{-1}$ ⇒ $Hg_{2} ^{+2}$

and then combine them together to form

$Mg + 2 Hg^{+2} + 2e$ ⇒ $Mg^{+2} + 2e ^{-} + 2 Hg^{+1} + 2 e^{-}$

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>

8 0

3 years ago